Let, (a + ib)2 = -15 - 8i
Now using, (a + b)2 = a2 + b2 + 2ab
⇒ a2 + (bi)2 + 2abi = -15 - 8i
Since i2 = -1
⇒ a2 - b2 + 2abi = -15 - 8i
Now, separating real and complex parts, we get
⇒ a2 - b2 = -15…………..eq.1
⇒ 2ab = -8…….. eq.2
⇒ a = - 4/b
Now, using the value of a in eq.1, we get
⇒ \((-\frac{4}{b})^2\) – b2 = -15
⇒ 16 – b4 = -15b2
⇒ b4 - 15b2 - 16 = 0
Simplify and get the value of b2, we get,
⇒ b2 = 16 or b2 = -1
As b is real no. so, b2 = 16
b = 4 or b = - 4
Therefore, a = -1 or a = 1
Hence the square root of the complex no. is -1 + 4i and 1 - 4i.